Given this function: $$f(z)=\frac{1}{(1-\cosh z)^2}$$
I must determine:
1) Principal part of the Laurent expansion at $z=0$
2) Radius of convergence of the analytic part.
1) I know that $z=0$ is a pole of order 4, since $$\lim_{z \to 0}z^4f(z)=4=c_{-4}$$ Wich is the coefficient of $\frac{1}{z^4}$. $$\frac{4}{z^4}+\frac{c_{-3}}{z^3}+\frac{c_{-2}}{z^2}+\frac{c_{-1}}{z}$$ But to get the other three coefficients I have to solve much more complicated expressions: $$c_{-k}=\frac{1}{(k-1)!}\lim_{z \to 0}\frac{d^{k-1}}{dz^{k-1}}[z^kf(z)]$$ Am I missing something? Is there a simpler way?
2) I think that the radius is $R=0$: $$f(z)=\left(\frac{1}{1-\cosh z}\right)^2=\left(\sum_{k=0}^\infty (\cosh z)^k\right)^2$$ So it must be $|\cosh z|<1$ wich is never true.
Simpler way: write $\cosh$ as a series expansion: $$\cosh(x)=1+\frac {x^2}{2!}+\frac{x^4}{4!}+\cdots\\\begin{align}\frac1{1-\cosh x}&=-\frac2{x^2}\frac{1}{1+\frac1{12}x^2+\frac1{360}x^4\cdots}\\&=-\frac2{x^2}\left(1-\frac1{12}x^2-\frac1{360}x^4+\frac1{144}x^4+O(x^6)\right)\end{align}$$Squaring this, $$\frac1{(1-\cosh x)^2}=\frac4{x^4}\left(1-\frac16x^2+\frac1{144}x^4-\frac2{360^2}x^4+\cdots\right)$$
The radius of convergence can be derived as follows:
$$\frac1{(1-\cosh z)^2}=\frac1{1-(\cosh z - \cosh^2z)}=\sum_n (\cosh z-\cosh^2z)^n$$This converges when $$|\cosh z-\cosh^2z|<1$$ You can then try to solve this. I believe the solution to this is something like $|z|<\left|\cos^{-1}\left(\frac{\sqrt3}2-\frac i2\right)\right|$